Problem: Solve for $x$ : $3x^2 + 12x - 15 = 0$
Dividing both sides by $3$ gives: $ x^2 + {4}x {-5} = 0 $ The coefficient on the $x$ term is $4$ and the constant term is $-5$ , so we need to find two numbers that add up to $4$ and multiply to $-5$ The two numbers $5$ and $-1$ satisfy both conditions: $ {5} + {-1} = {4} $ $ {5} \times {-1} = {-5} $ $(x + {5}) (x {-1}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x + 5) (x -1) = 0$ $x + 5 = 0$ or $x - 1 = 0$ Thus, $x = -5$ and $x = 1$ are the solutions.